# Tag Info

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This is not an answer, and don't take it as such. From a theoretical point of view, you have $A=\sum_{k=1}^{n}\lambda_k P_k$ where the $P_k$ are orthogonal projections. If you start iterating in $A$, or perform various functions of $A$, the problem is that $f(A)=\sum_{k=1}^{n}f(\lambda_k)P_k$. You can get down below the level of a projection. You can form ...

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Idea for an indirect proof: Suposse unit ball is compact. Cover it by $\cup_{x\in B_1 (0)} B_{\frac{1}{2}} (x)$. By compactness, there exist points finitely many points $a_i, i=1,\dots,n$ such that balls of radius $\frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.

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If it is infinite dimensional, we can find an infinite orthonormal basis, $(e_n)_{n \in \Bbb{N}}$. Notice $||e_j|| = 1$ and $||e_i - e_j|| = \sqrt{2}$ for $i \neq j$. (This is a very standard construction.)

1

Yes, but you have to verify this. Suppose $x=\sum f(v)v\in\operatorname{span}\beta$, where $v\in\beta$ and the sum is finite. Show that $\Vert Tx\Vert^2=\Vert x\Vert^2= \sum|f(v)|^2$ (use orthonormality of $\beta$). Thus, if $Tx=0$, we must have $x=0$, which means that $T$ is injective.

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Note: There is nothing about completeness of $\mathcal{H}$ needed to carry out of the following steps. Because $P_n$ is monotone, then $(P_nx,x)$ is monotone in $n$ for each fixed $x$, and is bounded above by $(x,x)$, which forces convergence of $\lim_n(P_n x,x)$ for all $x$. Then, using polarization, the following expression must also have a limit in $n$ ...

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Example, $$A = \frac{1}{i}\frac{d}{dx}$$ on the domain $\mathcal{D}(A)$ of absolutely continuous functions $f \in L^2[0,1]$ for which $f' \in L^2[0,1]$ and $f(0)=0$. Then $A^*$ is the same as $A$ except that the condition $f(0)=0$ is replaced by $f(1)=0$. Then $A^{\star\star}=A$ because $A$ is closed and densely-defined. However, ...

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Consider the two cases of $V_n$ being monotone increasing and monotone decreasing in $n$. In the first case you have $P_n P_{n+1}=P_n$ and in the second you have $P_n P_{n+1}=P_{n+1}$. At any rate you have either $P_n(P_{n+1}- P_{n})=0$ or $P_{n+1}(P_{n+1}-P_{n})=0$. First consider $V_n$ is increasing. So $(P_{n+1}-P_n)(z)$ is in $V_n^\perp$. This means ...

1

Assume $V_n \subseteq V_{n+1}$ for all $n$. Then $P_{n+1}P_n=P_n$. Using adjoint, $P_nP_{n+1}=P_n$ must also hold because $P_k^*=P_k$ for all $k$. Therefore, for all $x$, $$(P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x).$$

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In any infinite-dimensional Hilbert space $\mathcal H$, an orthonormal basis is not a basis in the algebraic sense. Suppose not: let $B$ be such an orthonormal basis. Let $b_1, b_2, \ldots$ be a sequence of distinct members of $B$ and $x = \sum_{j=1}^\infty b_j/j$. If $x = \sum_{b \in B} c_b b$ (where only finitely many $c_b \ne 0$), then $c_b = \langle b, ... 1 Yes, it is fine if you interpret$\langle D^2 f(x) , e_n \rangle$as the bilinear form $$(y,z) \mapsto \langle D^2 f(x)[y,z], e_n \rangle.$$ As you already said, this follows simply from the chain rule and the linearity of$L_n$. 2 If$s \in S$, then$\mathcal{N}_s = \{ x \in H : \langle x,s \rangle = 0 \}$is the null space of a continuous linear functional, which is the inverse image of$\{0\}$under this continuous linear functional. Hence$\mathcal{N}_s$is closed, as is the intersection $$S^{\perp} = \bigcap_{s\in S}\mathcal{N}_s.$$ 1 Let$T$be the Fréchet derivative of$f$at$e\in E$. Then $$\frac{|f_n(e+h)-f_n(e)-\langle Th,e_n\rangle|}{\|h\|} =\frac{|\langle f(e+h),e_n\rangle-\langle f(e),e_n\rangle-\langle Th,e_n\rangle}{\|h\|} =\frac{|\langle f(e+h)- f(e)- Th,e_n\rangle}{\|h\|} \leq\frac{\| f(e+h)- f(e)- Th\|}{\|h\|}\to0.$$ So the derivative of$f_n$is$h\longmapsto \langle ...

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Yes, this is true. Indeed, let $t_n$ be a sequence in $S^{\perp}$ and let $t$ be its limit. We want to show that $t \in S^{\perp}$. By the continuity of the inner product, we have that, for all $s \in S$ $$\langle t, s \rangle = \lim_{n \to \infty} \langle t_n, s \rangle=0$$ so that $t \in S^{\perp}$. The equality $(S^{\perp})^{\perp}=S$ only holds if ...

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It's true that $S^\perp$ is closed. This is trivial from the definition. This does not imply that $S$ is closed, because in general $S\ne S^{\perp\perp}$. In fact $S=S^{\perp\perp}$ if and only if $S$ is a closed subspace of $H$; in general $S^{\perp\perp}$ is the closure of the span of $S$.

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It's basically the same, it's mainly a matter of implicit identifications. Namely, in a Hilbert space $H$ there is a natural isomorphism $H\to H^*$ (which is the bra-ket duality), so $H\otimes H\simeq H\otimes H^*$. Now there is a map $H\otimes H^*\to L(H)$ which corresponds to what you call the outer product. So for two kets $|\psi\rangle$ and ...

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The first term contains a typo, it should read $B(x,1/2)$ instead of $B(x + 1/2)$. Moreover, $F + B(0,1/2)$ is the Minkowski sum, which is defined via $$F + B(0,1/2) = \{x + y \colon x \in F, y \in B(0,1/2)\}.$$ Hence, this equality is essentially the definition of the Minkowski sum.

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No it's not. For instance, on $\mathbb{C}^n$, take any matrix $M\in M_n(\mathbb{C})$ that is not hermitian, and put $(X,Y)\mapsto X^t M \overline{Y}$.

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This seems false. Take $U=H$, $Z=Id$, and identify $L(H,\mathbb{R})$ with $H$. Take and orthogonal base $e_1,e_2\ldots,$ and define $f(e_1)=e_2$, $f(e_i)=0$ for $i>1$, extend by linearity. Then $$\langle e_2,f(e_1)\rangle=1\neq0\langle e_1,f(e_2)\rangle.$$

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It seems that the uniqueness part has already been settled. In your argument regarding the existence part, you have proved that if the sequence $(x_n)_{n=1}^\infty$ you have constructed converges, then its limit belongs to $C$. But so far you have only that $\|x_n\| \rightarrow s$. It remains to be shown that $(x_n)_{n=1}^\infty$ does converge. To this end, ...

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There are two cases: $0\in C$ and $0\notin C$. In the first case the minimum norm is $=0$, and $x_0=0$ is the unique element with this property. If $x_0$ with minimum norm has already been found (and your question indicates that this is already known to you) assume $x_1\neq x_0$ has the same norm and it is an element of $C$. It is easy to see that $x_1$ and ...

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It's true that the $\ell^p$ and $\ell^q$, $p\ne q$, are not Lipschitz homeomorphic. (And changing a norm to an equivalent one does not change this.) This fact is not something that can be proved with bare hands, unless one of $p,q$ is $\infty$, when separability makes the distinction clear. Chapter 7 of Geometric Nonlinear Functional Analysis by Benyamini ...

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The inequality does not hold in general. Let $$T=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q_1=\begin{bmatrix}0&0\\0&1\end{bmatrix}, \ \ Q_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Note that $$... 1 The hint is to expand the inequality you have (for which you'll have to square), and to remember that V_n is a vector space. Assuming the space is real, if you expand from \|y_{n+1}\|\leq\|y-y_{n+1}\| you get$$\tag{1} \langle y,y_{n+1}\rangle\leq \frac12\|y\|^2,\ \ y\in V_n. $$Now, if we take y=y_k for some k\leq n,$$ \langle ...

0

Suppose we're in the unit disc $\mathbb D$ for simplicity. Let $\sum_{n=0}^{\infty}a_nz^n$ be the Taylor series of $f$ in $\mathbb D.$ Using the orthogonality of the exponenetials, we see $$\int_{\mathbb D}|f|^2\, dA = \int_0^1 \int_0^{2\pi} |f(re^{it})|^2\, dt \, r\, dr = \int_0^1\int_0^{2\pi}|\sum_{n=0}^{\infty}a_nr^ne^{int}|^2\, dt\, r\, dr$$ $$= ... 2 This is what in finite dimension is called the gradient of f, if it exists (and which you may call gradient in this case as well). It's the same idea as in the finite dimensional case. In order for this to work you need a natural isomorphism between the vector space and it's dual (which you usually don't have but) which is induced by the scalar product in ... 1 If e_1,\cdots,e_k is an orthonormal basis then \langle e_i,e_j\rangle is 1 when i=j and 0 when i\ne j. Consider the case with k=4 and m=3. We have$$\lambda_1\langle e_1,e_3\rangle+\lambda_2\langle e_2,e_3\rangle+\lambda_3\langle e_3,e_3\rangle+\lambda_4\langle e_4,e_3\rangle $$... 2 For an orthonormal basis, (e_n,e_m) = \delta_{nm}. \delta_{nm} = 1 when n=m and 0 when n\neq m. What happens when you plug this into your sum? 3 The trick is to show the contrapositive. If an operator T has infinite-dimensional and closed image, then it is not compact. Indeed, by restriction we get a bijective bounded linear operator T:(\ker T)^\perp\to\text{Im}\,T. By the open mapping theorem, T maps open sets to open sets. So the image of the unit ball is open, and this is not a compact ... 0 A^\perp = \operatorname{span}(A)^\perp = \overline{\operatorname{span}(A)}^\perp = \overline{A}^\perp and A^\perp is closed. This is clear from the following observations: x is orthogonal to A iff it is orthogonal to the span of A, by linearity of inner products. By continuity of inner products, if x_n \rightarrow x and all x_n obey \langle ... 0 Recall that the perp map is inclusion reversing. So suppose x \in (A^{\perp})^{\perp} is not in \mathrm{span}(A). Then (\mathrm{span}(A))^{\perp} = A^{\perp} is not contained in x^{\perp}, so there is some element of y \in A^{\perp} that is not orthogonal to x. 1 You are implicitly assuming that A_0 is injective. Let y\in H. Then there exists x\in D(A_0) with y=A_0x. Then$$ \langle A_0^{-1}y,y\rangle=\langle x,A_0x\rangle\geq0. $$Thus A_0^{-1} is positive, and so there exists an orthonormal basis \{e_n\} of eigenvectors. Since A_0^{-1} is compact, its eigenvalues \lambda_n satisfy ... 1 Yes, it is enough. Because A_0 is positive, symmetric and surjective, then A_0 is densely-defined, injective and selfadjoint. Therefore, A_0^{-1} is compact, selfadjoint with trivial null space. So A_{0}^{-1} has an orthnormal basis of eigenfunctions \{e_n \} with corresponding eigenvalue sequence of positive numbers$$ \lambda_1 \ge ...

1

Yes. Since $Z=|PTP|=(PTPTP)^{1/2}$, it is a limit of polynomials of the form $PX_nP$, and so $PZP=Z$. The spectral projections of an operator always belong to the von Neumann algebra generated by the operator. If $\mathcal M=W^*(|PTP|)=\{|PTP|\}''$, then $E^{|PTP|}(\Delta)\in\mathcal M$ for any Borel set $\Delta$. From the first paragraph we now that ...

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They key observation is that if $\lambda_n=0$ then $Qe_n=0$, and that $$\ker Q=\overline{\text{span}}\,\{e_n:\ \lambda_n=0\},\ \ \ \ (\ker Q)^\perp=\overline{\text{span}}\,\{e_n:\ \lambda_n>0\}.$$ To check this, let $x\in U$. We can write, since $\{e_n\}$ is an orthonormal basis, $$x=\sum_n\alpha_n\,e_n.$$ Then $$... 1 You have a linear bijection$$ Q^{1/2} : U \rightarrow U_0 $$Q^{1/2} is an isometric isomorphism because, by definition, Q^{1/2} is surjective, and is injective because \lambda_n > 0 for all n, and$$ \|Q^{1/2}y\|_{U_0}=\|Q^{-1/2}Q^{1/2}y\|_{U}=\|y\|_{U}. $$That also implies that \{ f_n=Q^{1/2}e_n \} is an orthonormal ... 1 A) Have a look at https://en.wikipedia.org/wiki/Complete_metric_space ex: \mathbb{R} is the completion of \mathbb{Q} for the natural distance. B) Here, your scalar product defines a norm, so a distance. For this distance, the completion of \mathcal{C}^{\infty}(S) is L^2(S). To be more precise, any Cauchy sequence of \mathcal{C}^{\infty}(S) cannot ... 1 The closed graph theorem indeed suggests itself: Suppose that f_n\to f and \phi f_n\to g in norm. We must show that then g=\phi f. This follows from$$ g(x)=\langle K_x, g\rangle = \lim\, \langle K_x, \phi f_n\rangle =\phi(x) \lim\, \langle K_x, f_n\rangle =\phi(x)f(x) . $$1 First notice that \overline{\mathrm{ran}(A-\lambda)}=\ker(A-\lambda)^\perp for \lambda\in\mathbb{R}. Thus, it suffices to show that \ker(A-\lambda)=\{0\} and that \mathrm{ran}(A-\lambda) is closed. Let \lambda<0. We have$$ \|(A-\lambda)u\|\|u\|\geq\langle (A-\lambda) u,u\rangle\geq -\lambda\|u\|^2 $$for all u\in D(A). Thus, ... 2 Note first that we may assume that all T_k are proper isometries; because if one of them, say T_1, is a unitary, we get \sum_{k=2}^NT_kT_k^*=0, which forces T_k=0 for all k\geq2. Since \sigma(T_k^*T_k)=\sigma(I)=\{1\}, using that \sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\} we deduce that \sigma(T_kT_k^*)=\{0,1\} (the zero has to be there ... 1 Define T on \ell^2 by$$T((x_1,x_2,\dots))=(0,x_1,x_2/2,x_3/3,\dots).$$2 Let \mathcal{H}=L^2[0,1]. Let \mathcal{D} be the subspace of polynomials on [0,1]. Let \mathcal{D}' be the subspace generated by \{ \sin(n\pi x) \}_{n=1}^{\infty}. Both are dense, and they have nothing in common except the 0 vector. 0 What about subspaces \mathbb Q and \mathbb R\setminus \mathbb Q in \mathbb R? 1 If \mu is sigma-finite, then there are disjoint Borel subsets \{ A_j \}_{j=1}^{\infty} of finite \mu-measure such that \bigcup_j A_j=\Omega. Let f_j = T1_{A_j}. Then$$ 1_{A_k}f_j = 1_{A_k}T1_{A_j}=T1_{A_k\cap A_j} =0 ,\;\;\; k \ne j,\\ 1_{A_j}f_j = f_j. $$So each function f_j is supported in A_j. Let f be the a.e. unique ... 6 Let \{a_n\} be a sequence of positive real numbers that increases to 1, with the property that the sequence of products$$ a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4,\ \ldots$$converges to a positive value. It's not hard to write down a specific example. Let \cal H = \ell^2(\mathbf R) and define T : \cal H \to \cal H by$$T(x_1,x_2,x_3,\ldots) = (0, ...

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Note that $\langle Tw, Tw\rangle=\langle w, w\rangle$. Taking $w=u-v$ gives $\langle T(u-v),T(u-v)\rangle=\langle(u-v),(u-v)\rangle\implies \left| u-v\right|^2=\left|Tu-Tv\right|^2$

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The $b_k$s must be bounded. Clearly this is a sufficient condition. It is also a necessary one: if $|b_k|$ is unbounded, then there is a subsequence $b_{k_n}$ such that $|b_{k_n}|>n$, and then we can choose $a_{k_n} = 1/b_{k_n}$ and keep the other $a_i$s zero. Then the $a_{k_n}$ are clearly square summable but $b_{k_n}a_{k_n}=1$ is not.

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For (b)$\Rightarrow$(a), assume $\left\Vert y_{n}\right\Vert <M$. For $\epsilon>0$, choose $x_{F}\in span\left\{ x_{k}\right\}$ s.t. $\left\Vert x-x_{F}\right\Vert <\epsilon/M$. Then \begin{eqnarray*} \left|\left\langle x,y_{n}\right\rangle \right| & \leq & \left|\left\langle x-x_{F},y_{n}\right\rangle \right|+\left|\left\langle ...

3

To complement Math1000's answer, the path of working with the equality $(P+Q-PQ)^2=P+Q-PQ$ cannot lead to a proof (i.e., the argument does not work for idempotents alone). Let $$P=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Then $PQ=Q$, $QP=P$, and $$P+Q-PQ=P=(P+Q-PQ)^2.$$

3

If $P+Q-PQ$ is a projection, then \begin{align} P+Q-PQ &= (P+Q-PQ)^*\\ &= P^* + Q^* - (PQ)^*\\ &= P^* + Q^* - Q^*P^*\\ &= P + Q - QP, \end{align} from which it follows that $PQ=QP$.

1

Second(*) inequality, follows from, $K=\bigcup\limits_{n\in\mathbb{N}_0}(\mathfrak{P}^{-j}+\mathfrak{p}^{-j}u(n))$. For third (**), use Parseval's identity.

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